where method results in a different outputs when using the inplace kwarg

[finete]

I am getting different results when using the where method with inplace=False vs inplace=True.

consider the following dataframe:

df = pd.DataFrame({'a':[1,None,3]})
df
    a
0   1.0
1   Nan
2   3.0

when using the where method with inplace=False (the default), i get the following result:

df.where(pd.notnull(df), None)
    a
0   1.0
1   None
2   3.0

with a being converted to be of type object.

however if i do this:

df.where(pd.notnull(df), None, inplace=True)
df
    a
0   1.0
1   Nan
2   3.0

a is of type float64

pandas version: 0.23.3 , numpy version: 1.14.5

reported because of my question on SO

[TomAugspurger]

I suppose the issue is that None cannot be stored in a float64 array, so if you want to place it then the array needs to be coerced to object.

It'd be helpful if someone could see how we values coercion with inplace=True in other places. I could see this raising if the other requires coercing self and inplace=True.

[finete]

I think the issue is specifically with None, because doing this:

df.where(pd.notnull(df), 'st', inplace=True)

does change a to be of type object (and of course changes the Nan to 'st').

maybe it's a misshadeling of a default argument somewhere down the line.

[mroeschke]

I think we consistently coerce None to NaN here now with inplace True/False (which I think is the expected behavior). Could use a test

In [6]: df = pd.DataFrame({'a':[1,None,3]})

In [7]: df.where(pd.notnull(df), None)
Out[7]:
     a
0  1.0
1  NaN
2  3.0

In [8]: df.where(pd.notnull(df), None, inplace=True)
   ...: df
Out[8]:
     a
0  1.0
1  NaN
2  3.0